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Each of the 25 balls in a certain box is either red, blue or [#permalink]
13 Aug 2010, 02:03

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A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

60% (02:14) correct
39% (01:28) wrong based on 204 sessions

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

(1) The probability that the ball will both be white and have an even number painted on it is 0

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

You can solve this problem in another way. Transform probability into actual numbers and draw the table.

Given:

Attachment:

1.JPG [ 8.64 KiB | Viewed 4587 times ]

So we are asked to calculate \frac{a+b-c}{25} (we are subtracting c not to count twice even balls which are white).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> c=0 --> \frac{a+b}{25}=?. Not sufficient.

Attachment:

4.JPG [ 8.7 KiB | Viewed 4590 times ]

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \frac{white}{25}-\frac{even}{25}=0.2 --> white-even=25*0.2=5 --> a-b=5 --> b=a-5 --> \frac{a+a-5-c}{25}=?. Not sufficient.

Attachment:

2.JPG [ 8.68 KiB | Viewed 4586 times ]

(1)+(2) c=0 and b=a-5 --> \frac{a+a-5+0}{25}=\frac{2a-5}{25}. Not sufficient.

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

how did you get this? P(WorE)=2P(E)+0.2 _________________

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

how did you get this? P(WorE)=2P(E)+0.2

From (1) P(WorE)=P(W)+P(E)-0 --> P(WorE)=P(W)+P(E); From (2) P(W)-P(E)=0.2 --> P(W)=P(E)+0.2;

Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 04:48

Expert's post

morya003 wrote:

Hi But can we not say that 12 out of the 25 balls were even ? If we can then we already get the answer with only B !.

Thanks & Regards

Well since OA is E then apparently we can not.

Also how did you get 12? Anyway even if we knew from (2) that # of even balls is 12 the answer still wouldn't be B since we would need # of balls which are both white and have an even number painted on them, so in this case answer would C. Please refer to: each-of-the-25-balls-in-a-certain-box-is-either-red-blue-or-99044.html#p763481 _________________

Re: Each of the 25 balls in a certain box is either red, blue or [#permalink]
03 Mar 2012, 06:32

Dont know why I have already put 1000 Dollars behind GMAT - lolz

I calculated 12 even numbers as follows - first 10 balls can be painted with 5 even numbers viz - 2,4,6,8,10 likewise for next 10 - 2,4,6,8,10 and next 5 - 2,4,

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

I found this method simple although it might look complicated. Took me exactly 2 minutes.
_________________

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

The interesting part about this explanation is particularly helpful expression "multiple values are possible for P(W) and P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)." I did not think of that at all (I just thought, well this is a minus probabilities and we need plus, so not sufficient- I am not even sure I understood any relevance of the 2nd option), when I finally thought of the problem in the same way. Now, may be it is quite late here and my brain refuses to come up with something , but are there instances in which multiple values are not possible and hence the answer would be B? Sort of a "what if" on this problem...
_________________

There are times when I do not mind kudos...I do enjoy giving some for help

Each of the 25 balls in acertain box is red, blue or white and has a number from 1 to 10 painted on it. If one ball is selected at random from the box , what is the probability that the ball selected will either be white or have an even number painted on it ?

a. the probability that the ball selected will be white and an even number painted on it is 0 b.The probability that the ball will be white minus an even number painted on it is 0.2
_________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

Each of the 25 balls in acertain box is red, blue or white and has a number from 1 to 10 painted on it. If one ball is selected at random from the box , what is the probability that the ball selected will either be white or have an even number painted on it ?

a. the probability that the ball selected will be white and an even number painted on it is 0 b.The probability that the ball will be white minus an even number painted on it is 0.2

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E)(0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2) Are these values only possible? If they are please explain why. Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?
_________________

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - P(W); Probability ball: even - P(E); Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)-0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)-P(E)=0.2, multiple values are possible for P(W) and P(E)(0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE).

(1)+(2) P(W&E)=0 and P(W)-P(E)=0.2 --> P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4 (10 even balls) then P(WorE)=1BUT if P(E)=0.2 (5 even balls) then P(WorE)=0.6. Not sufficient.

Answer: E.

Hope it's clear.

You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2) Are these values only possible? If they are please explain why. Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?

solution:

White balls = w Red = R Blue = b Total ball = 25 Sum total of even numbered ball = E

We have to evaluate = w/25 + (E)/25

From st(1) , we only know there is no white ball which contains even number. Even we still don’t know about red and blue balls have how many even numbered balls in them. So all are in mystery and doubt.

From st(2), w/25 – E/25 = 0.2 , again all unknown.

From both statement, we can assume several different things, I mean double case. So both are insufficient. Answer is (E)
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